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Taking the derivative of position to get velocity

  1. Feb 12, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem is hopefully attached, I had to take a screen shot.

    2. Relevant equations
    I understand the process of taking the derivative of position to get velocity.
    *refer to derivative rules.... for example r(x)=2x^2-6x+8 therefore r`(x)=4x-6

    3. The attempt at a solution
    I am stuck with what to do with the "1/s"

    The correct answer is 2.8m/s

    Thank you so much in advance.
     

    Attached Files:

  2. jcsd
  3. Feb 12, 2016 #2

    haruspex

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    I think the 1/s terms are just a confusing way of showing that the constant (0.74) has units of s-1.
     
  4. Feb 13, 2016 #3
    I may be wrong but here is what I think - To take cosine, sine etc. you need to have an angle. Clearly it is changing in the equation -as shown- with time. say at time equals to one second - (notice that second in denominator and that of time are going to vanish with each other) - the angle is equals to 0.74 (degrees/ rads/ sterdians etc.) But I need to know what is - m- and also speed is not constant - you must have been asked to find speed(*read velocity) at a particular time.
     
  5. Feb 13, 2016 #4

    haruspex

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    No, the .74 is a rate, radians per second, but radians are considered dimensionless, so it can be written as just s-1.
     
  6. Feb 13, 2016 #5
    It's the unit. Notice the m is also there, standing for "meters". But if you take the derivative you'll get an expression that still depends on "t", so I don't see how one could end up with 2.8 m/s. Velocity is a vector quantity, so, since the answer doesn't include any unit vectors, I suppose the problem asks for the speed. But we would need a given instant of time to get the numeric solution.
     
  7. Feb 13, 2016 #6

    HallsofIvy

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    There's the beginning of your problem! Velocity is the derivative with respect to time
    Assuming r(t) has units of "meters" then the derivative, which is the limit of the "difference quotient" has units of "meters per time" as velocity should.

     
  8. Feb 13, 2016 #7

    haruspex

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    i see no evidence that hsb was trying to do otherwise, nor for any issue over units.
     
  9. Feb 13, 2016 #8
    What I'm not seeing is at which point in time you're supposed to calculate the velocity... The velocity changes with time, and without knowing when to 'measure' it, we can't find a numerical answer, just the formula
     
  10. Feb 13, 2016 #9
    Thank you everyone for your help.
    When I take the derivative of the position vector I get:

    (-sin(0.74t)0.74)+(cos(0.74t)0.74)
    Then say at time 1.5, where the answer should equal 2.8m/s I get

    (-sin(0.74x1.5)0.74)+(cos(0.74x1.5)0.74)

    which equals
    -0.014+0.7397
    =0.7257

    Where am I messing up?
     
  11. Feb 13, 2016 #10
    You are messing up in two ways. You are omitting the 3 m and the 6 m coefficients. And the i and j represent unit vectors in the x and y directions. So both the displacement and the velocity have x and y components. To get the magnitude of the velocity, you need to consider the vector sum of the two components (using the Pythagorean theorem).
     
  12. Feb 13, 2016 #11
    Thank you Chester. I'm STILL stuck though ahah :(

    new attempt:

    (3)(-sin(0.74t)0.74)+(0)(cos(0.74t)+(6)(cos(0.74t)0.74)+(0)(sin0.74t)
    =(3)(-sin(0.74t)0.74)+(6)(cos(0.74t)0.74)
    =(3)(-sin(0.74x1.5)0.74)+(6)(cos(0.74x1.5)0.74)
    =-0.04+4.44

    then, velocity= sqrt(-0.04^2+4.44^2)

    I am getting 4.4....not 2.8...

    I am so sorry for the trouble ahah
     
  13. Feb 13, 2016 #12
    I get -1.988 i + 1.974 j. That gives 2.8 m/s for v. Are you sure you used radians, and not degrees?
     
  14. Feb 13, 2016 #13
    oh man...how embarrassing ahaha, thats what I was doing. :( Thank you so very much for your help! This is quite clear now. :) Have a great day!! :D Thank you!!!
     
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